SOLVING SYSTEM OF LINEAR AND QUADRATIC EQUATIONS FOR SAT

Problem 1 :

y = rx2 + s

y = -2

In the system of equations above, r and s are constants. For which of the following values of r and s does the system have exactly two real solutions ?

A)  r = -2, s = -1           B)  r = -1, s = -2

C)  r = 2, s = - 2           D)  r = 3, s = 1

Solution :

y = rx2 + s ----(1)

y = -2----(2)

Applying (2) in (1), we get

-2 = rx2 + s

Option A :

When r = -2, s = -1

-2 = -2x2 + (-1)

-2 + 1 = -2x2

-1 = -2x2

1/2 = x2

x = ±√(1/2)

Option B :

When r = -1, s = -2

-1 = -2x2 + (-2)

-1 + 2 = -2x2

1 = -2x2

-1/2 = x2

Not defined.

Hence option A is correct.

Problem 2 :

x2 - y2 = 48

x + y = 12

If (x, y) is the solution to the system of equations above, what is the value of xy ?

A)  28     B)  32     C)  45     D) 64

Solution :

x2 - y2 = 48 ----(1)

x + y = 12 ----(2)

y = 12 - x

x2 - y2 = 48

x2 - (12 - x)2 = 48

x2 - (144 - 24x + x2) = 48

x2 - 144 + 24x - x2 = 48

24x = 48 + 144

24x = 192

x = 192/24

x = 8

Applying the value of x in (2), we get 

y = 12 - 8

y = 4

xy = 8(4) = 32

So, the answer is 32.

Problem 3 :

y = -1.5

y = x2 + 8x + a

In the system of equations a is a positive a constant. The system has exactly one distinct real solution. What is the value of a ?

Solution :

y = -1.5 ---(1)

y = x2 + 8x + a ---(2)

Applying the value of y in (2), we get

-1.5 = x2 + 8x + a

x2 + 8x + a + 1.5 = 0

Finding nature of roots :

a = 1, b = 8 and c = a + 1.5

b2 - 4ac = 0

82 - 4(1)(a + 1.5) = 0

64 - 4 (a + 1.5) = 0

-4a - 6 = -64

-4a = -64 + 6

-4a = -58

a = 29/2

So, the value of a is 29/2.

Problem 4 :

y = 2x2 - 21x + 64

y = 3x + a

In the system of equation, a is a constant. The graphs of the equation in the system intersect at exactly one point (x,y) in the xy plane. What is the value of xy ?

Solution :

y = 2x2 - 21x + 64 ----(1)

y = 3x + a ----(2)

Applying (2) in (1), we get

3x + a = 2x2 - 21x + 64

2x2 - 21x - 3x + 64 - a = 0

2x2 - 24x + 64 - a = 0

Since it has a point of intersection at (x, y), the above quadratic equation will have real roots.

Finding nature of roots :

a = 2, b = -24 and c = 64 - a

b2 - 4ac = 0

(-24)2 - 4(2)(64 - a) = 0

576 - 8(64-a) = 0

576 - 512 + 8a = 0

64 + 8a = 0

8a = -64

a = -64/8

a = -8

y = 3x - 8

Applying the value of y,

3x - 8 = 2x2 - 21x + 64

2x2 - 21x + 64 - 3x + 8 = 0

2x2 - 24x + 72 = 0

x2 - 12x + 36 = 0

(x - 6)(x - 6) = 0

x = 6 and x = 6

y = 3(6) - 8

y = 10

xy = 6(10) ==> 60

So, the value of xy is 60.

Problem 5 :

x2 + 2xy + y2 = 25

x - y = 7

If (x, y) is a solution to the system of equations above, what is one possible value of x ?

Solution :

x2 + 2xy + y2 = 25 ---(1)

x - y = 7 ----(2)

x = 7 + y

Applying the value of x in (1), we get

(7 + y)2 + 2xy + y2 = 25

49 + 14y + y+ 2xy + y2 = 25

2y2+ 14y + 2xy + 49 - 25 = 0

2y2+ 14y + 2xy + 24 = 0

It cannot be solved.

Trying in another way :

(a + b)2 = a2 + 2ab + b2

(a - b)2 = a2 - 2ab + b2

a2 + b2 = (a - b)2 + 2ab

x2 + y2 = (x - y)2 + 2xy

Replacing the above in (1), we get

x2 + y2+ 2xy = 25

(x - y)2 + 2xy + 2xy = 25

72 + 4xy = 25

4xy = 25 - 49

4xy = -24

xy = -24/4

xy = -6

So, the value of xy is -6.

Problem 6 :

(1/2) x2 - 3y2 = 55

x = -4y 

If (x, y) is a solution to the system of equations above, what is the value of y2 ?

Solution :

(1/2) x2 - 3y2 = 55 -----(1)

x = -4y  -----(2)

Applying the value of x from (2) in (1)

(1/2) (-4y)2 - 3y2 = 55

(1/2) (16y)2 - 3y2 = 55

8y2- 3y2 = 55

5y2 = 55

y2 = 55/5

y2 = 11

So, the value of y2 is 11.

Problem 7 :

x - y + 2 = 0

(x + 2)2 - 7(x + 2) + 25 = 4y - 5

If (x, y) is a solution to the system of equations above, what is one possible value of x ?

Solution :

x - y + 2 = 0  ---(1)

(x + 2)2 - 7(x + 2) + 25 = 4y - 5 ----(2)

From (1),

y = x + 2

Applying the value of y in (2), we get

y2 - 7y + 25 = 4y - 5

y2 - 7y - 4y + 25 + 5 = 0

y2 - 11y + 30 = 0

(y - 6)(y - 5) = 0

y = 6 and y = 5

Applying y = 6, we get x = 6 - 2 ==> 4

Applying y = 6, we get x = 5 - 2 ==> 3

So, the values of x are 4 and 3.

Problem 8 :

f(x) = x2 - 3x

g(x) = 2x + 14

The functions f and g are defined above. For how many values of k is it true that f(k) = g(k) ?

A)  one    B)  None    C)  Two    D) more than two

Solution :

f(x) = x2 - 3x

g(x) = 2x + 14

f(k) = g(k)

k2 - 3k = 2k + 14

k2 - 3k - 2k - 14 = 0

k2 - 5k - 14 = 0

(k - 7) (k + 2) = 0

k = -2 and k = 7

So, it will have two solutions.

Problem 9 :

If (x + y)2 - (x - y)2 = 60 and x and y are positive integers, which of the following could be the value of x + y?

A)  6     B)  8     C)  10    D)  12

Solution :

(x + y)2 - (x - y)2 = 60

x2 + y2 + 2xy - (x2 + y2 - 2xy) = 60

x2 + y2 + 2xy - x2 - y2 + 2xy = 60

4xy = 60

xy = 60/4

xy = 15

The possible values of x and y are

x = 3, y = 5 or x = 1 and y = 15

x + y = 3 + 5 ==> 8

(or)

x + y = 1 + 15 = 16

So, option B is correct.

Problem 10 :

y = 3x - 1

y = (x + 1)2

The system of equations above has how many solutions ?

A)  0     B)  1     C)  2    D) infinitely many

Solution :

y = 3x - 1  ----(1)

y = (x + 1)2   ----(2)

(1) = (2)

3x - 1 = (x + 1)

3x - 1 = x2 + 2x + 1

x2 + 2x + 1 - 3x + 1 = 0

x2 - x + 2 = 0

(x + 2)(x - 1) = 0

x = -2 and x = 1

So, the system has two solutions, option C is correct.

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